Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k - 10}{k + 2} \div \dfrac{4k - 40}{k^2 - 4k - 12} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k - 10}{k + 2} \times \dfrac{k^2 - 4k - 12}{4k - 40} $ First factor the quadratic. $q = \dfrac{k - 10}{k + 2} \times \dfrac{(k + 2)(k - 6)}{4k - 40} $ Then factor out any other terms. $q = \dfrac{k - 10}{k + 2} \times \dfrac{(k + 2)(k - 6)}{4(k - 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k - 10) \times (k + 2)(k - 6) } { (k + 2) \times 4(k - 10) } $ $q = \dfrac{ (k - 10)(k + 2)(k - 6)}{ 4(k + 2)(k - 10)} $ Notice that $(k - 10)$ and $(k + 2)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(k - 10)}(k + 2)(k - 6)}{ 4\cancel{(k + 2)}(k - 10)} $ We are dividing by $k + 2$ , so $k + 2 \neq 0$ Therefore, $k \neq -2$ $q = \dfrac{ \cancel{(k - 10)}\cancel{(k + 2)}(k - 6)}{ 4\cancel{(k + 2)}\cancel{(k - 10)}} $ We are dividing by $k - 10$ , so $k - 10 \neq 0$ Therefore, $k \neq 10$ $q = \dfrac{k - 6}{4} ; \space k \neq -2 ; \space k \neq 10 $